Question

If \(\quad 10 \Sigma_{i=1}(x i-8)=9\) and \({ }^{10} \sum_{i=1}(x i-8)^{2}=45\) then standard deviation of \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \ldots \ldots \ldots \mathrm{x}_{10}\) is (a) \(19.2\) (b) \(12.92\) (c) \(1.82\) (d) \(1.92\)

Short answer

The standard deviation of the data points is approximately \(1.92\).

Step-by-step solution

Find the sum of all the data points

We have \(\displaystyle 10 \sum_{i=1}^{10}(x_i - 8) = 9\). This can be rewritten as \(\displaystyle \sum_{i=1}^{10}(x_i - 8) = \frac{9}{10}\). We can distribute the summation and get: \[\displaystyle \sum_{i=1}^{10}x_i - \sum_{i=1}^{10}8 = \frac{9}{10}\] Now, since each term in the second summation is 8, and there are 10 terms, the second summation simplifies to 80: \[\displaystyle \sum_{i=1}^{10}x_i - 80 = \frac{9}{10}\] From which we can find the sum of all \(x_i\): \[\displaystyle \sum_{i=1}^{10}x_i = 80 + \frac{9}{10} = 80.9\]

Find the mean

To find the mean, we simply divide the sum of all the data points by the number of data points (10 in this case): \[ \bar{x} = \frac{\sum_{i=1}^{10}x_i}{10} = \frac{80.9}{10} = 8.09\]

Calculate the variance

We are given \(\displaystyle { }^{10}\sum_{i=1}(x_i-8)^2 = 45\). To find the variance, we first rewrite the given expression as \(\displaystyle \sum_{i=1}^{10}(x_i-8)^2 = \frac{45}{10}\). Then, we substitute the mean we found earlier into the expression: \[\displaystyle \sum_{i=1}^{10}(x_i - 8.09)^2 = \frac{45}{10}\] First, we find the sum of the squared differences from the mean before calculating the variance: \[\displaystyle \sum_{i=1}^{10}(x_i - 8.09)^2 = \sum_{i=1}^{10}[(x_i - 8)^2 - (8.09 - 8)^2]\] \[\displaystyle \sum_{i=1}^{10}[(x_i - 8)^2 - 0.0081] = \frac{45}{10}\] Now, let us find the variance, which is the average of the squared differences from the mean: \[\text{Var}(x) = \frac{1}{10} \sum_{i=1}^{10}(x_i - 8.09)^2 = \frac{1}{10} \left[ \sum_{i=1}^{10}[(x_i - 8)^2 - 0.0081] \right] = \frac{45}{100}\]

Calculate the standard deviation

The standard deviation is simply the positive square root of the variance: \[\text{SD}(x) = \sqrt{\text{Var}(x)} = \sqrt{\frac{45}{100}} = \frac{\sqrt{45}}{10}\] This is equal to approximately \(1.92\). Therefore, the answer is (d) 1.92.

Key concepts

Summation

In mathematics, summation is an operation that combines a sequence of numbers into a single sum, using a Greek letter sigma (\(\Sigma\)) notation. In the context of our problem, summation was used to find the total of specific repeated calculations on data points. For example, \(\sum_{i=1}^{10}(x_i - 8)\) represents the sum of each is data point \(x_i\) from 1 to 10, minus 8. To calculate this, you:

• Subtract 8 from each of the data points \(x_i\).
• Add up all these adjusted terms together.

In the problem, we learned that \(\sum_{i=1}^{10}(x_i - 8) = 0.9\), representing the cumulative difference from 8, essential for further calculations in determining variance and standard deviation.

Variance

Variance is a crucial statistical concept that measures how much a set of numbers is spread out. It is calculated by averaging the squared differences between each data point and the mean of the dataset. This gives us an insight into the variability of the data.In this problem:

• The variance is calculated using \(\sum_{i=1}^{10}(x_i-8)^2\).
• The mean \(\bar{x} = 8.09\) was subtracted from each data point.
• Then, each of these differences was squared and summed up over all data points.

The step involved finding \(\sum_{i=1}^{10}[(x_i - 8)^2 - 0.0081]\) to refine the variance with respect to the actual mean. Finally, dividing this by 10 gave us the variance, indicating the average spread of the data points.

Mean

Mean, often referred to as the average, is a measure of central tendency of a set of numbers. It provides a single value that summarizes all the data, giving you an idea of the typical data point.To find the mean in this problem:

• First, find the total sum of all data points, which was \(80.9\).
• Divide this by the number of data points (10 in this case).

Thus, the mean \(\bar{x}\) came out to be \(8.09\). This calculated mean was then used in further steps to find the variance by questioning how each data point deviates from this central value.

JEE Maths

JEE Maths is a component of India's Joint Entrance Examination, essential for students aiming to pursue engineering. The mathematics section is known for its challenging problems that test deep understanding of concepts like standard deviation, variance, and mean. This mathematical exploration requires:

• Strong grasp of algebra and statistics concepts.
• Ability to apply logical steps in problem-solving as shown in this standardized problem about standard deviation.

JEE Maths problems, such as the one solved here, often require using concise calculations and applying theoretical knowledge, pushing students to think critically and understand deeply, preparing them for complex engineering studies.